Let $D \subset \mathbb C$ open, $f: D \to C$ a holomorphic function and $a \in D.$
a) Use
$D \subset \mathbb C open$
$\gamma$ rectifiable and closed in $D$ such that $n(\gamma,0) = 0, \forall w \in \mathbb C$ \ $D$. If $a \in D \setminus \{\gamma \} $ then $f^{k}(a) n(\gamma,a)=\frac{k!}{2 \pi i} \int_{\gamma}\frac {f(z)}{(z-a)^{k+1}} dz.$
to prove that if $a \in D, m \in \mathbb N$ and $z = a$ is a zero order $m$ of $f$, then $f(a) = … = f^{(m -1)}(a) = 0$ and $f^{(m)}(a) \neq 0$.
b) Use power series expansion to prove the converse of the previous item.
I do not know what to do.I can't manipulate it so it doesn't stay $z=a$ then
a) Knowing that $a$ is a zero of multiplicity $m$.Then there exists an analytic function $g : B(a, R) \to C$ such that $f(z) = (z − a)^mg(z)$ where $g(a) \ne 0$. Then, $h(z) = (z − a)^{m−1}g(z)$ has a zero of multiplicity $m − 1$ at $a$. Inductively, we assume that $h^{(m−2)}(a) = ... = h(a) = 0$ and $h^{(m−1)}(a) \ne 0$.$f^{(z)} = (z − a)h(z)$. So $f^{(i)}(z) = (z − a)h^{(i)}(z) + \sum_{i−1}^ {j=0} h^{(j)}(z)$. Then we see that $f^{(m−1)}(a) = ... = f (a) = 0$ and $f^{(m)}(a) \ne 0$.(i) Conversely, suppose $f^{(m−1)}(a) = ... = f (a) = 0$ and $f^{(m)}(a) \ne 0$. Let $a$ be a zero of multiplicity $k$. Then $f^{(k)}(a) \ne 0$, hence $k \ge m$, but $f^{(i)}(a) = 0$ for $i \lt k$ by (i). This implies that $k \le m$. So $k = m$.
b)$f(z)= \sum_{n=0}^{\infty}c_n(z-a)^n = 0$?
Thank's.
Clearly if $i\leq m-1$, then by the formula which you were supposed to use $$f^{(i)}(a)\eta(\gamma, a) =\frac{1}{2\pi i} \int\limits_{\gamma}\frac{(z-a)^mg(z) dz}{(z-a)^{i+1}} =\frac{1}{2\pi i} \int\limits_{\gamma}(z-a)^{m-i-1}g(z) dz $$ Since $m-i-1\geq 0$, for all $i$ with $0\leq i\leq m-1$, by Cauchy's theorem the last integral must be zero as $(z-a)^{m-i-1}g(z)$ is analytic in the given domain for non negative $i$ less than or equal to $m-1$. The same formula gives that $$f^{(m)}(a)\eta(\gamma, a) = g(a)\neq 0.$$ Hence $f^{(m)}(a)\neq 0$. This proves the first part.
For the converse part, $f$ has got a power series around $a$ given by $$f(z) = \sum\limits_{n=0}^{\infty} a_n(z-a)^n.$$ Here $$a_n= \frac{f^{(n)}(a)}{n!}.$$ By converse assumption, $a_n=0$ for all $n$, $0\leq n\leq m-1$. Hence $$f(z)= (z-a)^m \sum\limits_{n=m}^{\infty}a_n(z-a)^{n-m}.$$ This is of the form $$f(z) = (z-a)^mg(z),$$ Where $g(a) = a_m =\frac{f^{(m)}(a)}{m!}\neq 0.$